Let \(V_{i}=\lbrace f\,:\,G_{i}\rightarrow \mathbb {C}\rbrace \)  and for any element \(s\in G_{i}\) , define the \(s\) -adjecancy \(T_{s}\,:\,V_{i}\rightarrow V_{i}\) , \(T_{s}f(g)=f(gs)\) , and for any multiset \(S\)  of \(G_{i}\) , define the \(S\) -adjecancy operator \(T_{S}\,:\,V_{i}\rightarrow V_{i}\) , \(T_{S}=\sum _{s\in S}T_{s}\) .
Note that for any two multisets \(S,S^{\prime }\)  of \(G_{i}\) , \(T_{S\cup S^{\prime }}=T_{S}+T_{S^{\prime }}-T_{S\cap S^{\prime }}\)  and \(T_{S}T_{S^{\prime }}=T_{SS^{\prime }}\) , where \(SS^{\prime }=\lbrace ss^{\prime }\,:\,s\in S,s^{\prime }\in S^{\prime }\rbrace \)  counted with multiplicities.
Therefore \(T_{A_{i}^{\prime }}=T_{b}T_{B_{i}}+T_{B_{i}}T_{b}-2I\) , where \(b=\phi _{i}(b_{0})\) .
Let \(f\in V_{i}\)  be such that \(f\perp 1_{G_{i}}\) , i.e. \(\sum _{g\in G_{i}}f(g)=0\) .
Note that for any \(s\in G_{i}\) , then \(T_{s}f\perp 1_{G_{i}}\)  and \(\Vert T_{s}f\Vert =\Vert f\Vert \) .
By [1]}, we have \(\Vert T_{B_{i}}f\Vert \leqslant 2\sqrt{q}\Vert f\Vert \)  for any \(f\perp 1_{G_{i}}\) .
Then
\(\Vert T_{A_{i}^{\prime }}f\Vert \leqslant \Vert T_{b}T_{B_{i}}f\Vert +\Vert T_{B_{i}}T_{b}f\Vert +2\Vert f\Vert \leqslant \Vert T_{B_{i}}f\Vert +\Vert T_{B_{i}}(T_{b}f)\Vert +2\Vert f\Vert \) 
\(\leqslant \left(2\sqrt{q}-1\right)\Vert f\Vert +\left(2\sqrt{q}-1\right)\Vert f\Vert +2\Vert f\Vert =4\sqrt{q}\Vert f\Vert \leqslant 3\sqrt{2q-1}\Vert f\Vert \) 
